高等電機機械

學號： MA020211學生：周士平

題目 5.12• Consider the motor of Problem 5.10.• a. Compute the filed current required when the motor is

operating at rated voltage, 4200kW input power factor leading. Account for saturation inder load by the method described in the paragraph relating to Eq. 5.29.

• b. In addition to the data given in Problem 5.10, additional points on the open-circuit characteristic are given below:

• If the circuit breaker supplying the motor of part (a) is tripped, leaving the motor suddenly open-circuited, estimate the value of the motor terminal voltage following the trip.

Field current, A 200 250 300 350Line voltage, V 4250 4580 4820 5000

題目 5.12 Ans(a)

•The total power is

• and

•

kVAkW

pf

PS 4828

87.0

4200

05.29670aI 81.4038.0 jZ s

AAFNLI f 306)3/4160

4349(

題目 5.12 Ans(b)

If the machine speed remains constant and the field current is not reduced, the terminal voltage will increase to the value corresponding to 306 A of field current the open-circuit saturation characteristic. Interpolating the given data shows that this corresponds to a value of around 4850 V line-to-line.

題目 5.14

Loss data for the motor of Problem 5.10 are as follows：

Open-circuit core loss at 4160V = 37kWfriction and windage loss = 46kWfield-winding resistance at 75 =0.279Ω

Compute the output power and efficiency when the motor is operating at rated input power, unity power factor, and rated voltage. Assume the field-winding to be operating At a temperature of 125

題目 5.14 AnsAt rated power, unity power factor, the armature current will be Ia =5000 kW/(√3 4160 V) = 694 A. The power dissipated in the armature winding will then equal Parm = 3× 6942 × 0.011 = 15.9 kW.The field current can be found from|Eaf | = |Va − ZsIa| = |4160√3 − ZsIa| = 2394 V, line-to-neutralIf = AFNL_(2394/4160/√3) = 238 AAt 125C, the field-winding resistance will be Rf = 0.279(234.5 + 125/234.5+ 75)= 0.324 Ω andhence the field-winding power dissipation will be Pfield = I2f Rf = 18.35 kW.The total loss will then be Ptot = Pcore + Parm + Pfriction/windage + Pfield = 120 kWHence the output power will equal 4882.75 kW and the efficiency will equal 4882.75/5000= 0.976 = 97.6%.

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